College Algebra, Chapter 7, 7.1, Section 7.1, Problem 28

The system of linear equations

$
\left\{
\begin{equation}
\begin{aligned}

10x+10y-20z =& 60
\\
15x+20y+30z =& -25
\\
-5x+30y-10z =& 45

\end{aligned}
\end{equation}
\right.
$
has a unique solution.

We can rewrite the system in simplest form


$
\left\{
\begin{equation}
\begin{aligned}

x+y-2z =& 6
&& \frac{1}{10} \times \text{Equation 1}
\\
\\
3x + 4y + 6z =& -5
&& \frac{1}{5} \times \text{Equation 2}
\\
\\
-x + 6y - 2z =& 9
&& \frac{1}{5} \times \text{Equation 3}

\end{aligned}
\end{equation}
\right.
$


We use Gauss-Jordan Elimination

Augmented Matrix

$\left[ \begin{array}{cccc}
1 & 1 & -2 & 6 \\
3 & 4 & 6 & -5 \\
-1 & 6 & -2 & 9
\end{array} \right]$

$\displaystyle R_3 + R_1 \to R_3$


$\left[ \begin{array}{cccc}
1 & 1 & -2 & 6 \\
3 & 4 & 6 & -5 \\
0 & 7 & -4 & 15
\end{array} \right]$

$R_2 - 3R_1 \to R_2$

$\left[ \begin{array}{cccc}
1 & 1 & -2 & 6 \\
0 & 1 & 12 & -23 \\
0 & 7 & -4 & 15
\end{array} \right]$


$R_3 - 7R_2 \to R_3$

$\left[ \begin{array}{cccc}
1 & 1 & -2 & 6 \\
0 & 1 & 12 & -23 \\
0 & 0 & -88 & 176
\end{array} \right]$


$\displaystyle \frac{-1}{88} R_3$

$\left[ \begin{array}{cccc}
1 & 1 & -2 & 6 \\
0 & 1 & 12 & -23 \\
0 & 0 & 1 & -2
\end{array} \right]$


$\displaystyle R_1 - R_2 \to R_1$

$\left[ \begin{array}{cccc}
1 & 0 & -14 & 29 \\
0 & 1 & 12 & -23 \\
0 & 0 & 1 & -2
\end{array} \right]$


$\displaystyle R_1 + 14R_3 \to R_1$

$\left[ \begin{array}{cccc}
1 & 0 & 0 & 1 \\
0 & 1 & 12 & -23 \\
0 & 0 & 1 & -2
\end{array} \right]$


$\displaystyle R_2 - 12R_3 \to R_2$

$\left[ \begin{array}{cccc}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & -2
\end{array} \right]$


We now have an equivalent matrix in reduced row-echelon form and the corresponding system of equations is


$
\left\{
\begin{equation}
\begin{aligned}

x =& 1
\\
y =& 1
\\
z =& -2

\end{aligned}
\end{equation}
\right.
$


Hence we immediately arrive at the solution $(1,1,-2)$.